If principal argument of z satisfying the inequalities - z -3- -2 and - z -6-3 i - - 2 -2 is - then tan -A-B-C- 21-8D- 8-15

The correct option is D 8/15|z 3|≤√(2) ; circle with centre (3, 0), radius =√(2), circle with centre (6, 3), radius =2√(2). Now, C_1C_2 = √(9+9)=3 √(2)=r_1+r ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Monotonically Increasing Functions

If principal ...

Question

If principal argument of z satisfying the inequalities $| z - 3 | \leq \sqrt{2} a n d | z - 6 - 3 i | \leq 2 \sqrt{2} i s θ$ , then tan $θ$ =

\frac{2}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{8}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{21}{7}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{8}{15}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{8}{15}$
$| z - 3 | \leq \sqrt{2}$ ; circle with centre (3, 0), radius $= \sqrt{2}$ , circle with centre (6, 3), radius $= 2 \sqrt{2}$ .

Now, $C_{1} C_{2} = \sqrt{9 + 9} = 3 \sqrt{2} = r_{1} + r_{2}$
$\Rightarrow$ $C_{1} a n d C_{2}$ touch externally
$\Rightarrow$ Their point of contact, P is the only point satisfying both inequalities
P divised $C_{1}$ (3, 0) and $C_{2}$ (6, 3) in ratio 1 : 2
$\Rightarrow$ Coordinates of P are (4, 1)
$\Rightarrow z = 4 + i \Rightarrow t a n θ = \frac{1}{4} \Rightarrow t a n 2 θ = \frac{8}{15}$

Suggest Corrections

Similar questions

Q. Let

z

be a complex number having the argument

θ, 0 < θ < π / 2

and satisfying the equality

| z - 3 i | = 3

. Then

cot θ - \frac{6}{z}

is equal to

Q. If principal argument of

z_{0}

satisfying

| z - 3 | \leq \sqrt{2}

and

arg (z - 5 i) = - \frac{π}{4}

simultaneously is

θ,

then the CORRECT statement(s) is/are

Q. If

z = \frac{- 2 (1 + 2 i)}{3 + i}

where

i = \sqrt{- 1}

, then the argument

θ (- π < θ \leq π)

z

Q. If

a r g (z^{1 / 3}) = \frac{1}{2} a r g (z^{2} + ¯ ¯ ¯ z z^{1 / 3})

, then find the value of

| z |

.
(Note :

z

has both real and imaginary components. If

θ

is the argument of

z

, then use

\frac{θ}{3}

as the argument of

z^{1 / 3}

and

0 < θ < \frac{π}{2}

)

Q. Let z satisfy

| z | = 1

and

z = 1 - ¯ z .

Statement-1: z is a real number.
Statement-2: principal argument of z is

\frac{π}{3} .

Join BYJU'S Learning Program

If principal argument of z satisfying the inequalities |z−3|≤√2 and |z−6−3i|≤2√2 is θ, then tan θ =

If principal argument of z satisfying the inequalities $| z - 3 | \leq \sqrt{2} a n d | z - 6 - 3 i | \leq 2 \sqrt{2} i s θ$ , then tan $θ$ =